Isothermal Heat Transfer

[VincentG]

Except that I don't consider "heat energy disappears" and "internal energy of the gas reduces" two separate. Operations.

Heat going in becomes "internal energy".

Presumably incorporated in some way into the expanding gas. Gas "expanding" is just another way of saying the gas molecules "increase kinetic energy".

The more energetic molecules impact the piston in some way transferring the energy to the piston which moves down the cylinder.

Basically all just transfers of molecular kinetic energy. From the molecules of the hot plate to the working fluid to the piston, like so many billiard balls.

At the end of the series of operations there is no"heat" that was supplied remaining in the working fluid. The working fluid only acted as an intermediary for transferring energy to the piston.

There is then no need to transfer "heat" to any "sink".

This is a fundamental issue that needs to be resolved here. With expansion towards BDC, the internal energy can remain constant while temperature lowers. I think you are suggesting that the internal energy of the gas itself is reduced as the piston expands towards BDC and that the heat energy transforms into mechanical work. Sorry If I'm being dense and not understanding you.

The inverse of this would be internal energy increasing while the gas is being compressed towards TDC.

This is exactly why I proposed this question.

Suppose an engine is in a perfectly insulated room. The engine is exactly 30 percent "efficient". We send 100 units of heat energy to a perfectly insulated engine running under full load to lift a mass within the room. How many units of heat will go on to heat the room?

[Tom Booth]

...

This is a fundamental issue that needs to be resolved here. With expansion towards BDC, the internal energy can remain constant while temperature lowers.

Not in my opinion.

Speaking generally, ("ideal gases") internal energy and kinetic energy are the same thing. The measure of kinetic energy/internal energy (in an ideal gas, or gases generally under most common conditions) is temperature.

You don't have constant internal energy (of an ideal gas) and lower (or higher) temperature.

I think you are suggesting that the internal energy of the gas itself is reduced as the piston expands towards BDC and that the heat energy transforms into mechanical work.

If the temperature goes down, the internal energy has been reduced. Temperature is the measure of the internal energy or the "kinetic energy". Same thing but different terms for different models.

Also the "heat energy" of the gas is also the "internal energy" or has become the internal energy. These are not two different things that can somehow be viewed separately, or treated separately.
Heat is the transfer of energy into the gas which energy then can be called "internal" but it is the same energy that was transfered in as "heat".

The inverse of this would be internal energy increasing while the gas is being compressed towards TDC.

Generally speaking, if work is being done on the gas.

IMO, if the gas outputs work during expansion very quickly (adiabatically, too fast for heat transfer from the surroundings) the gas may "contract" so rapidly the piston really does scarcely any work "compressing" the gas. ("A "non-ideal" gas behavior).

This is exactly why I proposed this question.

Suppose an engine is in a perfectly insulated room. The engine is exactly 30 percent "efficient". We send 100 units of heat energy to a perfectly insulated engine running under full load to lift a mass within the room. How many units of heat will go on to heat the room?

I would say that given the described conditions, virtually none.

First of all you say the engine is perfectly insulated, so the only way heat or energy gets out is through work. But the work is stored as potential energy, by lifting the mass.

If the engine is "only 30% efficient'" however, 70 units of heat are presumably trapped in the engine, not sure what you mean by "send". Can the heat go into the room rather than the engine? Also if the engine is "perfectly insulated" how does the heat get in? But I assume this is not a trick question so, likely the unused heat will cause the engine to overheat and quit running. Then I suppose the heat would eventually leak out, but you did say the engine is"perfectly insulated.

Also how does the heat get into a perfectly insulated room? Electric wires?

Frankly the question seems a bit no sensical.

[VincentG]

The engine is perfectly insulated so no heat is lost other than what is exhausted or rejected to the sink. The heat is supplied to the engine within the confines of an isolated chamber so as to not heat the room directly.

The only thing heating the room is what is exhausted/rejected to the sink.

[Tom Booth]

The engine is perfectly insulated so no heat is lost other than what is exhausted or rejected to the sink. The heat is supplied to the engine within the confines of an isolated chamber so as to not heat the room directly.

The only thing heating the room is what is exhausted/rejected to the sink.

In that case, I'd say 70 units of heat would be "rejected" to the room. The rest going out as "work" is tied up as potential energy in the lifted weight. Until, or unless it is dropped. Then the 30% will be returned as heat, sound, vibration...

That is all assuming the engine and room were in thermal equilibrium prior to the addition of the 100 units sent to the engine.

Also assuming we're talking about an otherwise "perfect" frictionless lossless (no heat conducted through the engine body rather than the working fluid) engine.

In other words, the room is initially the same temperature as the engine and in particular the cold plate before heat is added.

But first, before heat can be transfered out of the engine, the temperature of the cold plate will be increased by the "70 units" of "waste heat". Or perhaps that 70 will be divided over some time interval. Say 10 per cycle?

Whatever the case, some temperature rise at the cold plate would be a consequence and a necessity for heat to be transfered out to the room through that plate.

Some rise in temperature at the cold plate should be detectable if the engine efficiency is less than 100%

It might also be considered what happens if the weight is lifted outside the insulated room by one means or another, so that the "work" does not accumulate within the room.

[VincentG]

In that case, I'd say 70 units of heat would be "rejected" to the room. The rest going out as "work" is tied up as potential energy in the lifted weight.

Ok, now were getting somewhere.

Now how do we tie in the theory that there is a potential lowering of gas temperature and refrigeration effect? Does that counter act the 70 units of heat being rejected? Or does the refrigeration effect come out of the work attained from 30 units of heat?

[Tom Booth]

In that case, I'd say 70 units of heat would be "rejected" to the room. The rest going out as "work" is tied up as potential energy in the lifted weight.

Ok, now were getting somewhere.

Now how do we tie in the theory that there is a potential lowering of gas temperature and refrigeration effect? Does that counter act the 70 units of heat being rejected? Or does the refrigeration effect come out of the work attained from 30 units of heat?

If your asking me, conversion of heat into work is a "refrigeration effect" of a sort.

100 joules or whatever "heat unit" going in, 30 joules converted to work. The temperature of the working fluid was raised 100 joules then lowered 30.

Let's say the 100j raised the temperature from 300°k to 400°k heat conversion then "refrigerated" the gas back down to 370°k

Of course that is tied to efficiency. You've strictly limited that to 30%. Assuming you mean actual thermal efficiency not "Carnot" efficiency, which is essentially meaningless.

100% thermal efficiency would convert all the input heat to work, which might be used to run a generator to light lights 1000 yards away. Theoretically the work output would not effect the local environment as the ambient atmosphere is an "infinite" reservoir.

For an actual "refrigeration" effect, the thermal efficiency would need to be 100% plus.

That is lets say 100% of the heat is converted to work in expansion.

Some of that heat is converted into the form of velocity/momentum.

This velocity/momentum could theoretically be used to further expand the working fluid, bringing the temperature of the working fluid below ambient

This in fact appears to be not all that uncommon if we recall this graph:

Temperature_vs_angle.jpg (61.48 KiB) Viewed 2317 times

.A Stirling engine converts heat to work with a potential theoretical efficiency of 100% on the expansion stroke. Why could the "work" not be used for the purpose of refrigeration, by expanding the working fluid further?

[Tom Booth]

I mean, in principle. If you store "potential energy" in a bolder, by rolling it up a hill, it can then roll down. I don't think this is any different then driving out a piston "up hill" against atmospheric pressure. It can then return.

But what happens when the bolder reaches the bottom of the hill? Does it stop?

Actually you have only then reached a point where the work of pushing the bolder up the hill can be recovered, now stored as velocity and momentum in the bolder.

At the top of the hill it was only "potential energy". When it reaches the bottom of the hill that has been converted to kinetic energy.

Kind of like this demonstration. The wind up car rolls further back past where it started still carrying kinetic energy

https://youtu.be/begl90R_gRo?si=tKeK7KNVRkHc8Oui

It's only the Carnot myth that alleges that "heat" is somehow different from other types of energy conversion.

[Stroller]

The further the boulder makes it up the hill on the other side of the valley, due to it's momentum; the harder it was to push it up the hill on this side of the valley, due to its inertia.

[Tom Booth]

The further the boulder makes it up the hill on the other side of the valley, due to it's momentum; the harder it was to push it up the hill on this side of the valley, due to its inertia.

So?

[Tom Booth]

I mean, we are only talking about a piston returning to TDC after being pushed "up hill" against 1 atmosphere. Similar to lifting a weight against gravity or winding a spring.

If it is "harder" to lift a heavier weight, the equivalent is still stored as "potential energy" and returned when the weight hits the ground or when the spring unwinds.

I'm willing to acknowledge loses to friction and so forth. I'm not talking "perpetual motion" or "free energy" just getting back most of what you put in, or beating the "Carnot Limit" which insists 90% of the input heat just mysteriously "disappears" to the "cold reservoir" but for some reason just can't be detected actually going anywhere.

The piston in a Stirling engine (using the term "Stirling' loosely, to include thermal lag, thermoacoustic, "free piston" etc.) returns to TDC after being driven to BDC by the expanding gas against 1atm with no displacer to move the working fluid to a cold "sink", no flywheel and no effective "heat" outlet, just the 1atm that was displaced.

The whole notion that the piston can't return without "heat rejection" just doesn't really make sense, but more than that doesn't hold up experimentally.

[VincentG]

If your asking me, conversion of heat into work is a "refrigeration effect" of a sort.

I'm still working out exactly what I am asking so bear with me lol. And I mean thermal efficiency.

Air cools down when it expands, and heats when it's compressed. If the air cools down further than it would normally have just by expansion alone, then it will have a lower peak temperature when it is recompressed again. This would be the reduction of internal energy. Otherwise, if it just cools down as the adiabatic index says, the internal energy is constant, but now it's spread out and temperature is lower while the gas occupies a larger area.

If the air cools down further than the adiabatic index says it should(and internal energy is decreased), is that due to the gas performing work on the piston, or the piston performing work on the gas?

If it's due to the piston performing work on the gas, then it can only reduce power output, right?

If it's due to the gas performing work on the piston, does that increase or reduce the heat passing through the engine.

And if an engine is 100% thermally efficient, does that mean all the heat disappears as it's converted to work and the room is not heated at all? Or does(should) that mean all the heat is allowed to pass to the sink and heat the room while work is simultaneously extracted?

[VincentG]

I've been avoiding the rock analogy. But to me rolling the same rock up and down a hill seems pointless. How about we ask, if the expansion stroke rolls a rock uphill, can the return stroke roll another rock up the same hill, thus performing more useful work?

[Tom Booth]

If your asking me, conversion of heat into work is a "refrigeration effect" of a sort.

I'm still working out exactly what I am asking so bear with me lol. And I mean thermal efficiency.

Air cools down when it expands, and heats when it's compressed. If the air cools down further than it would normally have just by expansion alone, then it will have a lower peak temperature when it is recompressed again. This would be the reduction of internal energy. Otherwise, if it just cools down as the adiabatic index says, the internal energy is constant, but now it's spread out and temperature is lower while the gas occupies a larger area.

If the air cools down further than the adiabatic index says it should(and internal energy is decreased), is that due to the gas performing work on the piston, or the piston performing work on the gas?

The gas performing work on the piston.

Along with the additional work due to the piston driving the crankshaft that drives a generator or whatever else.

If it's due to the piston performing work on the gas, then it can only reduce power output, right?

If it's due to the gas performing work on the piston, does that increase or reduce the heat passing through the engine.

Reduce,

since the heat is converted to work.

And if an engine is 100% thermally efficient, does that mean all the heat disappears as it's converted to work and the room is not heated at all?

Yes

Or does(should) that mean all the heat is allowed to pass to the sink and heat the room while work is simultaneously extracted?

No. That is the old caloric theory.

I think maybe these questions are best answered by considering air-cycle refrigeration or air conditioning.

Most "normal" vapor compression air conditioning systems use a compressor to compress a gas to a higher pressure, (leaving out phase change. Let's just stick with a gas cycle.)

The gas is compressed which increases its temperature. While still under compression it is allowed to cool or actively cooled with a heat exchanger. Then the compressed gas is released through a valve and allowed to expand, that is, to escape from the compressed condition. This is called Joule-Thomson refrigeration. Just ordinary expansion through a valve.

Here is a very simple example:

https://youtu.be/2hYQtB4QkEY?si=qZC2D4L2wiSfi8W2

The gas gets pretty cold when it expands by this method, which as can be seen from the video, takes nothing more than an ordinary shop compressor.

But that is NOT "air cycle" refrigeration.

Air cycle refrigeration is often used for cryogenics. Serious cold.

Air cycle refrigeration can produce ultra cold temperatures. Way colder than the "snow" produced by that guys shop compressor.

The only difference really is that instead of just allowing the gas to expand into the open air like in the video, the air is used to drive an air motor on the way out of the air tank making it do additional work as it expands.

The air motor drives some load. Let's say, some fans, or a generator or even just a break that does nothing but provide resistance so that the gas does additional work driving the air motor and the load as it escapes from the tank.

So the gas doesn't escape from the tank through a valve and get cold, instead it escapes through an air motor making the gas do work as it expands and it gets much much colder.

The additional work done by the gas in such a way, causes the gas to cool down much more than it would otherwise, by just escaping from being under pressure. Doing the additional work driving the air motor converts additional heat into work so that the gas cools down to the cryogenic range.

Technically the "air motor" is called an expansion engine or expansion turbine, but people often get confused by this thinking the expansion engine does the work of expanding the gas. In actuality the gas expands doing the work to drive the expansion engine.

As I've related before, this can be done to some extent using just an ordinary compressor and any ordinary air motor.

But to be really effective the air motor would need to be insulated so that it doesn't absorb heat from the surroundings.

If the motor is not insulated the surrounding ambient heat will reduce the cooling effect. The ambient heat would heat the air and prevent it from getting really super intensely cold.

So returning to your last question:

And if an engine is 100% thermally efficient, does that mean all the heat disappears as it's converted to work and the room is not heated at all?

Yes. Assuming the work is not converted right back into heat in some way.

Or does(should) that mean all the heat is allowed to pass to the sink and heat the room while work is simultaneously extracted?

That is the Carnot theory. Some people still believe that is the case. Just like water can turn a turbine and the turbine turns a generator and the water does not "disappear". That is what Carnot believed when he wrote his book. That is the Caloric theory of heat that says heat is a fluid like water that runs down from hot to cold.

Heat, however, is actually just energy. When energy is transformed, it ceases to exist in its original form. So you cannot transform heat into work and have the heat go out to a sink and simultaneously go out as work. That would be doubling the energy to 200%

It would be like you walking out of a room through two different doors at the same time. You'd have to be in two places at the same time. Impossible.

100% going to work and 100% still going to a sink as heat. It's impossible. It's a violation of the conservation of energy. You can't just double your energy out of nothing.

When heat is transformed into work, the "heat" no longer exists.

The ENERGY exists.

Well does energy really exist? I don't know.

I have a hard time sometimes fathoming in what way "potential energy" is "stored" in a rock taken to the top of a hill. If you break the rock open, will you find this "potential energy" inside?

All I know is air-cycle air conditioning works. It is used on most aircraft. It works by making the gas do work.

The Claude method of liquifying gas works.

It works by making the gas do work as it expands in a cylinder, just like gas expanding and doing work as it expands in the cylinder of a Stirling engine. Not as high pressure, not cryogenic cooling, but same principle. When a gas expands and does work at the same time it cools down much much more than by just expansion alone.

[Tom Booth]

I've been avoiding the rock analogy. But to me rolling the same rock up and down a hill seems pointless. How about we ask, if the expansion stroke rolls a rock uphill, can the return stroke roll another rock up the same hill, thus performing more useful work?

Not exactly.

The point is, in doing work while expanding the gas (working fluid) gets cold due to converting heat into work.

The work is "useful work" that the engine does like running an electric generator. You can't get that work back.

But by doing work and loosing energy and getting cold, the gas contracts.

By contracting, this allows outside atmospheric pressure to do the work of returning the piston to TDC.

Kind of like the expanding gas pushes the "rock" up the hill.

Then atmospheric pressure gives it a shove so that it rolls back down.

Since atmospheric pressure is also doing work pushing the piston, now in the opposite direction, it should be possible to also extract some additional work on the return stroke.

If the gas was not provided with additional heat on the return stroke, the piston would be driven right into the inside of the engine.

In a lot of engines it seems like the return stroke is more forceful than the "power" stroke.

I think, taking the rock analogy, it is a slow struggle pushing the rock all the way up a hill, but when it rolls back down, watch out!!

I've often seen this fooling around with thermoacoustic engines, like this guy:

https://youtu.be/DyPxNNJQo9M?si=ZEXr3AcXtfCcqXzk

He says he had to install a spring to keep the piston from banging into the orifice on the return stroke.

I've had similar experiences where the piston in a "free piston" engine gradually got closer and closer to the inside end of the cylinder until it started actually banging into the inside end of tbe cylinder. Pretty weird to see the piston moving in that direction, toward the hot end, while I'm blasting that end with a propane torch.

[Stroller]

Since atmospheric pressure is also doing work pushing the piston, now in the opposite direction, it should be possible to also extract some additional work on the return stroke.

Only if some working fluid has been bled from the engine at working temperature near TDC. Otherwise, you're not going to get the partial vacuum at BDC needed for the ambient atmosphere to do any pushing on the return stroke.

But then the increased pressure on the power stroke has to overcome that negative pressure differential before it can start to do work accelerating the flywheel.

So there's no net gain to be had here. You're only changing the position of the PV cycle on the pressure axis, not changing the area within it.

I've had similar experiences where the piston in a "free piston" engine gradually got closer and closer to the inside end of the cylinder until it started actually banging into the inside end of tbe cylinder. Pretty weird to see the piston moving in that direction, toward the hot end, while I'm blasting that end with a propane torch.

I suspect the 'overshoot' will be occurring at both ends of the free piston travel as temperature, pressure differential and thus velocity increases.